I am working on a practice problem with p-iodotoluene and $\ce{CH3O-}$ in $\ce{CH3OH}$ at $\mathrm{25^{\circ}C}$. Another reaction is run which is identical except there is heat and pressure. Neither of these scenarios produce a reaction according to the answer key. What am I missing?

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$\begingroup$Would you mind describing what's on your answer key & some necessary details on 'except there is heat and pressure'?$\endgroup$
– MockingbirdFeb 23 '17 at 1:11

$\begingroup$That's all the question and answer key honestly say. I can say CH3)(-) is in CH3OH with p-iodotoluene and is heated and pressurized. Beyond that, I wasn't told anything else unfortunately.$\endgroup$
– Biomed BoyFeb 23 '17 at 1:15

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$\begingroup$This is not something we can answer in its present state (excluding those who have the ability of mind-reading).$\endgroup$
– JanFeb 23 '17 at 1:16

$\begingroup$I am confused by what you mean. How could I change the circumstances to to have enough of a question with clarity? It's still an intro course so I am not well-versed.$\endgroup$
– Biomed BoyFeb 23 '17 at 1:19

$\begingroup$@Ryan Wisth But what is 'another reaction'?$\endgroup$
– MockingbirdFeb 23 '17 at 1:19

1 Answer
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(This answer is made on assumptions made on your question; maybe it's wrong.)

p-iodotoluene is nucleophilic. So, in order to happen a reaction with p-iodotoluene, you need an electrophile. But $\ce{CH3-}$ is not an electrophile, rather a nucleophile.

But, you can get $\ce{CH3+}$ from methanol. But, methanol breaks into $\ce{CH3+}$ and $\ce{OH-}$ only in presence of a strong proton donor (i.e $\ce{CH3OH + H+<--> CH3+ + H2O}$) which is not present here, and I think it's very unlikely to have a strong proton donor in this alcoholic solution.

And the reaction

$$\ce{CH3^-X+ + CH3OH-> CH3+ + CH3- + XOH}$$

seems bogus to me as alcohol won't have that polarity to break a ionic bond.

So nothing seems plausible. Increase of heat and pressure doesn't change the scenario.